CSB349H1_08SPRING1_246 - Copy

CSB349H1_08SPRING1_246 - Copy - CSB349H1S Term Test 1...

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Unformatted text preview: CSB349H1S Term Test 1 Answer Key 1 1. The human gene VT is expressed in the brain and heart during early development, but in zebrafish there are two copies of this gene called VT-A and VT-B. In zebrafish, the VT-A gene is expressed only in the brain while the VT-B gene is expressed only in the heart. (a) Briefly describe how the duplication-degeneration-complementation (DDC) model of genomic evolution can explain the diverged expression of the two VT genes in zebrafish compared to the human VT expression pattern. (3 marks) Ancestral VT gene is duplicated in the teleost lineage after the divergence of mammalian (human) and teleost (zebrafish) lineages (1 mark) After the genes have duplicated, there is a loss of sequence conservation in the two separate regulatory regions (mutations accumulated differently over evolutionary time) (1 mark) (“degenerate” = 0.5 marks) Changes in regulatory sequences of the two genes leads to changes in gene expression (VT-A in brain and VT-B in heart), such that they compliment one another with regards to the ancestral gene expression pattern (could also say subfunctionalization) (1 mark) - co ID : 56 m 8 - r ID : dd 83 y. 67 st (b) What experiment in zebrafish embryos would you perform to prove that the upstream regulatory regions of VT-A and VT-B have evolved separate functions in controlling gene expression in zebrafish embryos in vivo? (Be sure to describe the basic steps in the experiment and the predicted results) (3 marks) 8 56 : tu de .c dy ud tb en y dd Bu nt om ad e lo wn Do - bu Engineer separate enhancer-reporter constructs using upstream regulatory regions from gene VT-A (0.5 marks) and gene VT-B (0.5 marks) as well as a reporter (e.g. EGFP) (0.5 marks) Generate trangenic zebrafish embryos for each construct (0.5 mark) Show that each enhancer can direct the expression of reporter to either brain (VT-A) or heart (VT-B) (0.5 marks each) - ID 67 st 83 Te .s ID : on to ud st o. Do w nl oa de r Engineer enhancer-reporter construct (0.5 marks) using human VT gene (0.5 marks) upstream regulatory region and reporter (0.5 marks) Generate zebrafish transgenics (0.5 marks) Expect to see reporter expression in brain AND heart (1 mark) ut or /u - nt ro to - ht tp :/ Uo fT St ud en t (c) Using zebrafish transgenic technology, describe how you would prove that the expression pattern of the human VT gene is likely to be most similar to the ancestral vertebrate expression pattern before DDC has taken place in the fish lineage. (Be sure to describe the basic steps in the experiment and the predicted results) (3 marks) 2. Promoter clearance is defined by the disassembly of most of the components of the pre-initiation complex and the synthesis of a short stretch of ~30 nucleotides of new mRNA by RNA Pol II, after which transcription is stalled. (a) What are the 2 main factors that cause RNA Pol II stalling during this phase of transcription? (0.5 marks each) - NELF (0.5 marks) Spt4/Spt5 complex or DSIF (0.5 marks) only NTEF (0.25 marks) capping (0 marks) This test is copyrighted by the uploader/owner. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. 2 (b) Briefly describe the two main steps that are required to overcome these negative factors and allow RNA Pol II to enter productive elongation. (2 mark) P-TEFb phosphorylates Spt4/Spt5 (0.5 marks), which causes the release of NELF (0.5 marks) P-TEFb phosphorylates S2 of CTD (0.5 marks), which helps recruit other RNA processing enzymes (e.g splicing factors) while RNA Pol II is elongating (0.5 marks) - (c) What experiment would you perform to prove that RNA Pol II has “escaped” the promoter of a given gene and is undergoing productive transcription elongation within cells? Be sure to state why you chose this experiment and describe the basic steps of the experimental protocol. State one possible control that you would include in your experiment (4 marks). Chromatin IP using RNA Pol II antibodies to detect occupancy of RNA Pol II (1 mark) Design ChIP primers for target region somewhere in the downstream coding region because RNA Pol II showed be in productive elongation (1 mark) - ChIP protocol: Fix cells; shearing DNA; IP with specific antibody & purify IP’s DNA; PCR with specific primers or perform microarray experiment to identify the target (0.25 each up to 1 mark) - Possible control experiments: PCR primers for promoter (expect to see RNAPII occupying promoter region even if it doesn’t escape); ChIP for a standard protein-DNA interaction you expect to see all the time; ChIP using non-specific IgG antibody (expect no PCR band); PCR with input sheared DNA (expect to see band with target primers) (1 mark) NOTE: Any marks you received for question 2C are included in your total final mark, but the exam is out of 36 marks due to the removal of question 2C from the test total. m co y. 67 dd 83 : ID 8 de .c dy ud tb y dd nt om bu r ad e lo wn Do 56 : ID st tu .s en ud 67 83 Te to Do w nl oa de r ID : on st o. ut or ro to /u :/ Uo fT St - nt en Helicase (0.5 marks) – uses ATP to unwind DNA duplex and expose template strand (0.5 marks) Kinase (0.5 marks) – phosphorylates S5 of CTD which helps to recruit RNA capping enzymes to the CTD ud - t Bu 3. Name two distinct enzymatic functions associated with the general transcription factor complex TFIIH (0.5 each). For each, briefly state its role in regulating transcription (0.5 each). 4. Briefly describe the 4 main steps that are important for transposition of non-replicative DNA transposons in the genome (4 marks). tp ID : 56 8 - ht st CSB349H1S Term Test 1 Answer Key - Transposase gene encoded by the transposon is transcribed/translated into a functional enzyme (1 mark) Transposase makes blunt-ended cuts in donor DNA (which releases the transposon DNA) and staggered (5’ overhang) cuts in target DNA (1 mark) Transposase ligates the transposon DNA to the 5’ single stranded ends of the target DNA (1 mark) Cellular DNA Pol extends 3’ cut ends and cellular ligase joins 3’ ends to transposon 5’ ends (generating direct repeats on either side of newly integrated transposon) (1 mark) This test is copyrighted by the uploader/owner. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. CSB349H1S Term Test 1 Answer Key 3 5. Answer the following 4 questions related to chromatin: (a) What is the fundamental unit of chromatin (0.5 mark)? Nucleosome (0.5 marks) - (b) Describe its core protein (2 marks) and DNA (1 mark) components. octamer protein complex composed of 2 copies of H2A, H2B, H3, H4 histones (0.25 marks for each protein and 1 mark for stating that there are 2 copies of each) 147 nucleotides of genomic DNA wrapped around ~1.7 turns of core proteins (1 mark) - (c) List three 3 common covalent protein modifications found on members of the core proteins (0.5 marks each). Acetylation (0.5 marks for each of 3 chosen from the list) Methylation Phosphorylation Ubiquitylation m co y. 67 st ID : 56 8 - ATP hydrolysis (1 mark) Nucleosome displacement (1 mark) nt de ud y dy Do .c wn lo om ad e - bu r ID : dd 83 (d) List 2 key features of Class II nucleosome remodeling complexes, which distinguish them from Class I remodeling complexes (2 marks). ht tp - 8 56 : ID 83 Te 67 st tu ID : on to .s tb en ud st o. nt This test is copyrighted by the uploader/owner. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. r de nl oa Do w ut or - Promoter histones containing H2A.Z are acetylated (0.5 marks) These modified histones further recruit class II remodeling complexes (e.g. SWI/SNF) (0.5 marks) Results in the removal of the H2A.Z histones from the DNA – called “histone eviction”(0.5 marks) Binding of mediator (0.5 marks) Further binding of class II remodeling complexes and histone eviction (0.5 marks) Formation of the Dnase I hypersensitive site (nucleosome free region) (0.5 marks) ro to - /u :/ Uo fT St ud en t Bu dd 6. The current model describing the mechanism for nucleosome-dependent transcription in yeast holds that transcriptional activation proceeds with the binding of sequence specific transcription factors (activators) to enhancers, which help to recruit chromatin modification complexes to the promoter. In point form, outline the key steps that occur before pre-initiation complex (PIC) assembly takes place in the region of the transcription start site. (3 marks) CSB349H1S Term Test 1 Answer Key 4 7. You are a cancer researcher studying tumorigenesis and are specifically interested in Li-Fraumeni syndrome – a condition characterized by a predisposition to the early onset of cancer. Recent work has demonstrated that mutations in the promoter of p53, a crucial tumor suppressor gene, are found in a large proportion of Li-Fraumeni individuals. (a) Outline a strategy by which you could determine which regions of the p53 promoter are required for gene expression. (2 marks) Deletion (linker scanning mutagenesis)/point mutation (1 mark) and transfection into cell culture for enhancer/reporter assays (1 mark) - EMSA using a radiolabeled p53 promoter sequences for the mutant and wild type promoters + protein extracts (1 mark) and an anti-NF1 antibody to show a supershift (1 mark) Alternative answers: ChIP, DnaseI assay (part marks only) m 67 - co st ID : 56 - r ID : dd 83 y. 8 (b) It is also known that the transcription factor NF1 binds as a monomer to a consensus-binding site in the p53 promoter. Briefly describe how you could determine whether a given point mutation in the p53 promoter disrupts direct NF1 binding? (2 marks) bu lo om ad e 8. Answer the following 3 questions related to repetitive DNA: nt de ud y dy Do .c wn (a) Provide example of a tandemly repeated DNA element (1 mark) and list two criteria that define this element (0.5 marks each). tp ht 8 56 : ID 83 Te 67 st tu : ID r de Do w nl oa ut or on to .s tb en ud st o. nt ro to /u Minisatellite DNA (1 mark) -clusters up to 20 kb in length with repeat units up to 25 bp (0.5) -ex. Telomere (0.5) :/ Uo fT St ud en t Bu dd Satellite DNA (1 mark): -non-coding sequences arranged in tandem arrays (0.5) -repeated millions of times in the genome (0.5) -constitute a substantial portion of the genome (0.5 -10’s to 100’s of bp in size (0.5) -more commonly contain AT rich sequences (0.5) -located in heterochromatin (0.5) Microsatellite DNA (1 mark) -clusters <150 bp in length w. repeat units of 2-6 bp (0.5) -highly variable (0.5) -located in euchromatin (0.5) -generated by replication slippage (0.5) -CA containing repeats are most common in the human genome (0.5) (b) Tandemly Repetitive DNA elements can be used as genetic markers to distinguish different genetic strains from one another, why? (0.5 mark) Repetitive elements can be polymorphic and therefore vary in size from one genetic strain to another. This test is copyrighted by the uploader/owner. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. CSB349H1S Term Test 1 Answer Key 5 (c) An aquarium contains a mixture of the following different genetic strains of fish: Homozygous Strain A, Homozygous Strain B and Heterozygous Strain A/B. Strain A fish can be identified by a 200 bp SSLP polymorphism and Strain B fish can be identified by a 300 bp SSLP polymorphism. Which column(s) in the gel below correspond to each of the three different genetic strains found in the aquarium? Ladder 1 2 3 4 5 6 1000bp co bu nt de dy Do .c wn lo om ad e r ID : 100 bp dd 83 y. 67 st ID : 56 m 8 500 bp 8 : ID 67 st ht This test is copyrighted by the uploader/owner. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. 83 : ID r de nl oa Do w ut or on nt ro to answer: 3 (0.5 marks) tp :/ /u fT St Strain A/B: Column #_________ answer: 5 (0.5 marks) Te to st o. ud en t Strain B: Column #__________ Uo answer: 1 (0.5 marks) .s en ud Bu Strain A: Column #__________ 56 tu ud tb dd y For each fish strain, list all of the appropriate columns: ...
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This note was uploaded on 12/03/2010 for the course CSB CSB349 taught by Professor V.tropepe,a.moses during the Fall '10 term at University of Toronto- Toronto.

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